Answer to Question #268057 in Mechanics | Relativity for JERRY

Question

Block 1 of mass 𝑚1 = 2 𝑘g is placed on block 2 of mass 𝑚2 = 3 𝑘g which is then placed on a table. A string connecting block 2 to a hanging mass 𝑀 passes over a pulley attached to one end of the table, as shown in Fig. 1. The mass and friction of the pulley are negligible. The coefficients of friction between block 1 and block 2 are 𝜇_1s = 0.2, 𝜇_1k = 0.1 and between block 2 and the tabletop are 𝜇_2s = 0.3, 𝜇_2k = 0.15

(a) (Marks: 3) Draw the complete free body diagram of the system and determine the largest value of 𝑀 for which the blocks can remain at rest.

(b) (Marks: 4) Now suppose that 𝑀 = 2.5 𝑘g which is large enough so that the hanging block descends and block 1 slips on block 2. Draw the complete free body diagram of the system and determine each of the followings.

i. The magnitude 𝑎1of the acceleration of block 1.

ii. The magnitude 𝑎2 of the acceleration of block 2.

Answer 1

a)

The maximum friction force on the blocks on the table:

$F_{fr2max}=\mu_{s2}N_2=\mu_{s2}(m_1+m_2)g$

which is balanced by the weight of the hanging mass:

$Mg=\mu_{s2}(m_1+m_2)g$

then:

$M=\mu_{s2}(m_1+m_2)=0.3g(2+3)=14.7$ kg

b)

i)

$F_{fr1}=\mu_{k1}m_1g=m_1a_1$

$a_1=\mu_{k1}g=0.1g=0.98$ m/s²

ii)

for the two blocks:

$Mg-T=Ma_2$

$T-F_{fr1}-F_{fr2}=m_2a_2$

$a_2=\frac{M-\mu_{k1}m_1-\mu_{k2}(m_1+m_2)}{M+m_2}g=\frac{2.5-0.1\cdot2-0.15(2+3)}{2.5+3}g=2.76$ m/s²